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2 years ago

What is the answer to this question?

Mathematics

2 answers:

steposvetlana [31]2 years ago

7 0

Answer:

48x - 40y +24

Step-by-step explanation:

Dimas [21]2 years ago

3 0

Answer:

D (48x - 40y +24)

Step-by-step explanation:

First, you need to distribute the 8 throughout the whole equation, so 8×6x=48x, 8×-5y=-40y, and 8×3=24. Right away, you can see that the answer is D.

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Solve. -7x=x^2+12 help

Nat2105 [25]

Hey There!

The answer to your problem is $x=-3$ or $x=-4$ both answers are correct!

Step 1: Subtract x²+12 from both sides.

−7x−(x²+12)=x²+12−(x²+12)

−x²−7x−12=0

Step 2: Factor left side of equation.

$(-x-3)(x+4)=0$

Step 3: Set factors equal to 0.

$-x-3=0$ or $x+4=0$

Get your answer:

$x=-3$ or $x=-4$

3 0

3 years ago

Please help. Thank you.

Marysya12 [62]

The answer would be 24pi inches.

You can find this by using the formula for a circumference and plugging in the value of the radius.

C = 2pi*r

C = 2pi * 12

C = 24pi

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2 years ago

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marusya05 [52]

Answer:

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Step-by-step explanation:

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2 years ago

Please help me find the area of this irregular shape.

ANTONII [103]

Answer:

23 Cm(2)

Step-by-step explanation:

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6 0

2 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

8 0

2 years ago